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SAT Math: Algebra — Complete Study Guide

Master linear equations, functions, systems, and inequalities for the Digital SAT. Covers all Algebra skills tested, with worked examples, strategies, and practice problems.

Last updated: February 10, 2026Estimated reading time: 2 hours

Introduction

The Algebra domain is the most important part of SAT Math. On the Digital SAT, Algebra questions test your ability to model real situations with linear relationships, manipulate equations accurately, and reason about constraints. If you can solve linear equations quickly and correctly, interpret slope and intercept in context, and set up systems from word problems, you will earn points in almost every SAT Math module.

You should expect about 13–15 Algebra questions out of 44 total Math questions, which is roughly 30% of the section. That is why Algebra is usually the #1 priority when building a study plan. Improving your Algebra accuracy by even a few questions can have a meaningful effect on your final scaled score.

Algebra appears in both Module 1 and Module 2. Module 1 includes straightforward skill checks (isolating variables, evaluating functions, reading slope). Module 2 can include more layered tasks, like modeling with constraints, systems written in context, and interpreting equivalent forms. In other words, Algebra is not just frequent; it is also foundational for higher-difficulty items.

On test day, Algebra rewards three habits:

  1. Translate words into equations before calculating.
  2. Keep your algebraic steps clean and reversible.
  3. Check whether your answer makes sense in context (units, sign, reasonableness).

Strategy tip: On SAT Math, most "hard" algebra questions are not about advanced theory. They are about avoiding small setup errors under time pressure.


1. Linear Equations in One Variable

Concept explanation

A linear equation in one variable is an equation where the variable has exponent 1 and appears in expressions like:

ax+b=cax + b = c

where aa, bb, and cc are constants and a0a \neq 0.

Your goal is to isolate xx using inverse operations. SAT questions in this category frequently test:

  • Distribution: correctly applying multiplication over parentheses.
  • Combining like terms: grouping xx-terms with xx-terms and constants with constants.
  • Fractions: clearing denominators with the least common denominator (LCD).
  • Special cases: recognizing when equations have no solution or infinitely many solutions.

A reliable approach:

  1. Expand parentheses first.
  2. Move variable terms to one side, constants to the other.
  3. Simplify completely.
  4. Check for special forms such as 0=00=0 (infinite solutions) or 0=50=5 (no solution).

For equations with fractions, multiply every term on both sides by the LCD. This removes denominators and reduces arithmetic mistakes.

Worked Example 1: Basic one-variable equation

Worked Example 1: Solve 3(2x - 5) + 4 = 2x + 7

We want to solve:

3(2x5)+4=2x+73(2x - 5) + 4 = 2x + 7
  1. Distribute 3 across the parentheses:

    6x15+4=2x+76x - 15 + 4 = 2x + 7
  2. Combine like terms on the left:

    6x11=2x+76x - 11 = 2x + 7
  3. Subtract 2x2x from both sides:

    4x11=74x - 11 = 7
  4. Add 11 to both sides:

    4x=184x = 18
  5. Divide by 4:

    x=184=92x = \frac{18}{4} = \frac{9}{2}

So the value of xx is:

92\boxed{\frac{9}{2}}

Quick check:

  • Left side: 3(2925)+4=3(95)+4=34+4=163(2\cdot\frac{9}{2}-5)+4 = 3(9-5)+4 = 3\cdot4+4=16
  • Right side: 292+7=9+7=162\cdot\frac{9}{2}+7 = 9+7 = 16

Both sides match, so the solution is correct.

Worked Example 2: Equation with fractions

Worked Example 2: Solve (x + 3)/4 - (2x - 1)/6 = 5/12

Solve:

x+342x16=512\frac{x+3}{4} - \frac{2x-1}{6} = \frac{5}{12}
  1. Find the LCD of 4, 6, and 12.

    • LCD is 12.
  2. Multiply every term by 12:

    12(x+34)12(2x16)=12(512)12\left(\frac{x+3}{4}\right) - 12\left(\frac{2x-1}{6}\right)=12\left(\frac{5}{12}\right)
  3. Simplify each term:

    3(x+3)2(2x1)=53(x+3) - 2(2x-1) = 5
  4. Distribute:

    3x+94x+2=53x+9 -4x+2 = 5
  5. Combine like terms:

    x+11=5-x + 11 = 5
  6. Subtract 11 from both sides:

    x=6-x = -6
  7. Multiply by 1-1:

    x=6x = 6

Answer:

6\boxed{6}

Quick check in the original equation:

6+342(6)16=94116=272212=512\frac{6+3}{4}-\frac{2(6)-1}{6}=\frac{9}{4}-\frac{11}{6}=\frac{27-22}{12}=\frac{5}{12}

Correct.

SAT Strategy Box

Strategy tip: If the equation is messy, rewrite each step on a new line. The SAT rewards organization as much as speed.


2. Linear Equations in Two Variables

Concept explanation

A linear equation in two variables describes a line on the coordinate plane. You should be fluent with the three common forms:

  1. Slope-intercept form:

    y=mx+by = mx + b

    where mm is slope and bb is the yy-intercept.

  2. Standard form:

    Ax+By=CAx + By = C
  3. Point-slope form:

    yy1=m(xx1)y-y_1 = m(x-x_1)

You must convert between forms quickly. For example, from standard form to slope-intercept form, solve for yy.

To find slope from two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2):

m=y2y1x2x1m = \frac{y_2-y_1}{x_2-x_1}

Line relationships:

  • Parallel lines have the same slope.
  • Perpendicular lines have slopes that are negative reciprocals (if m1m2=1m_1m_2=-1).

SAT context problems often hide these ideas inside words like "per hour," "each month," "starts with," or "initial fee." Those words usually map directly to slope and intercept.

Worked Example 3: Write equation from context

Worked Example 3: Plumber cost model

A plumber charges a $75 service fee plus $50 per hour.

  1. Write an equation for total cost CC in terms of hours hh.

  2. Find the cost for 3.5 hours.

  3. Identify slope and intercept:

    • Fixed fee (starting amount): 7575
    • Rate per hour: 5050
  4. Write linear model:

    C=50h+75C = 50h + 75
  5. Substitute h=3.5h=3.5:

    C=50(3.5)+75C = 50(3.5) + 75
  6. Compute:

    C=175+75=250C = 175 + 75 = 250

Final answers:

C=50h+75,C(3.5)=250\boxed{C = 50h + 75}, \qquad \boxed{C(3.5)=250}

So the cost for 3.5 hours is $250.

Worked Example 4: Interpreting slope and intercept in context

Worked Example 4: Population model interpretation

Given:

P=25,000+1,200tP = 25{,}000 + 1{,}200t

where tt is years after 2020.

Interpret 1,200 and 25,000.

  1. Compare to slope-intercept form y=mx+by=mx+b.

    • Here, slope m=1,200m=1{,}200.
    • Intercept b=25,000b=25{,}000.
  2. Interpret slope in context:

    • Population increases by 1,200 people per year.
  3. Interpret intercept in context:

    • At t=0t=0 (year 2020), the population is 25,000.

Answer:

  • 1,2001{,}200 is the yearly population change.
  • 25,00025{,}000 is the initial population in 2020.

Strategy tip: In context questions, plug in x=0x=0 (or t=0t=0, h=0h=0) to interpret the intercept immediately.


3. Linear Functions

Concept explanation

A linear function is often written as:

f(x)=mx+bf(x) = mx + b

This is the same line idea, written in function notation. SAT questions may ask you to:

  • Evaluate functions, such as finding f(3)f(3).
  • Solve for input when output is given, such as finding xx when f(x)=10f(x)=10.
  • Interpret rate of change from equations, tables, or graphs.
  • Determine whether a function is increasing or decreasing.

Key ideas:

  • If m>0m>0, the function is increasing.
  • If m<0m<0, the function is decreasing.
  • If m=0m=0, the function is constant.

From a table, rate of change is:

ΔyΔx\frac{\Delta y}{\Delta x}

For linear functions, this ratio stays constant between all pairs of points.

Domain and range reminders:

  • For an unrestricted linear function, domain is all real numbers and range is all real numbers.
  • In context problems (like time or quantity), domain may be restricted (e.g., t0t\ge 0).

Worked Example 5: Function evaluation

Worked Example 5: If f(x)=4x-7, find x when f(x)=f(3)+10

Given:

f(x)=4x7f(x)=4x-7

We need xx such that:

f(x)=f(3)+10f(x)=f(3)+10
  1. Compute f(3)f(3):

    f(3)=4(3)7=127=5f(3)=4(3)-7=12-7=5
  2. Compute right side:

    f(3)+10=5+10=15f(3)+10=5+10=15
  3. Set function equal to 15:

    4x7=154x-7=15
  4. Solve:

    4x=22x=224=1124x=22 \Rightarrow x=\frac{22}{4}=\frac{11}{2}

Answer:

112\boxed{\frac{11}{2}}

Worked Example 6: Rate of change from table

Worked Example 6: Find rate of change and function rule from a table

Given table:

xxf(x)f(x)
05
211
417
623
  1. Find rate of change:

    • From x=0x=0 to x=2x=2, Δx=2\Delta x=2, Δf=115=6\Delta f=11-5=6.
    • So slope is: m=ΔfΔx=62=3m=\frac{\Delta f}{\Delta x}=\frac{6}{2}=3
  2. Confirm with another interval:

    • From x=2x=2 to x=4x=4, Δf=1711=6\Delta f=17-11=6, Δx=2\Delta x=2, slope still 3.
    • Constant rate confirms linear function.
  3. Find intercept using x=0x=0 row:

    f(0)=5b=5f(0)=5 \Rightarrow b=5
  4. Write function rule:

    f(x)=3x+5f(x)=3x+5

Final answers:

  • Rate of change: 3\boxed{3}
  • Function: f(x)=3x+5\boxed{f(x)=3x+5}

Strategy tip: In a table, always test two intervals. If the slope changes, the function is not linear.


4. Systems of Linear Equations

Concept explanation

A system of linear equations asks for values that satisfy two linear equations at the same time. Geometrically, this is the intersection point of two lines.

Methods:

  1. Substitution: solve one equation for one variable, substitute into the other.
  2. Elimination: add/subtract equations to cancel one variable.

Possible outcomes:

  • One solution: lines intersect once (different slopes).
  • No solution: lines are parallel (same slope, different intercepts).
  • Infinitely many solutions: same line (equivalent equations).

For equations in slope-intercept form:

  • If m1m2m_1 \neq m_2: one solution.
  • If m1=m2m_1 = m_2 and b1b2b_1 \neq b_2: no solution.
  • If m1=m2m_1 = m_2 and b1=b2b_1 = b_2: infinitely many solutions.

System word problems usually involve two unknowns (items, prices, counts, rates). Define variables clearly before writing equations.

Worked Example 7: System by elimination

Worked Example 7: Solve 2x + 3y = 14 and 4x - y = 10

Solve:

{2x+3y=144xy=10\begin{cases} 2x+3y=14\\ 4x-y=10 \end{cases}
  1. Multiply the second equation by 3 so yy terms will cancel:

    12x3y=3012x-3y=30
  2. Add with the first equation:

    (2x+3y)+(12x3y)=14+30(2x+3y)+(12x-3y)=14+30 14x=4414x=44
  3. Solve for xx:

    x=4414=227x=\frac{44}{14}=\frac{22}{7}
  4. Substitute into 4xy=104x-y=10:

    4(227)y=104\left(\frac{22}{7}\right)-y=10 887y=10=707\frac{88}{7}-y=10=\frac{70}{7} y=707887=187-y=\frac{70}{7}-\frac{88}{7}=\frac{-18}{7} y=187y=\frac{18}{7}

Solution:

(227,187)\boxed{\left(\frac{22}{7},\frac{18}{7}\right)}

Quick check in first equation:

2227+3187=44+547=987=142\cdot\frac{22}{7}+3\cdot\frac{18}{7}=\frac{44+54}{7}=\frac{98}{7}=14

Correct.

Worked Example 8: System word problem

Worked Example 8: Notebooks and pens

A store sells notebooks for $3 and pens for $1.50. A student buys a total of 12 items and spends $27. How many of each did they buy?

  1. Define variables:

    • Let nn = number of notebooks.
    • Let pp = number of pens.
  2. Write equations:

    • Total items: n+p=12n+p=12
    • Total cost: 3n+1.5p=273n+1.5p=27
  3. Clear decimal in second equation (multiply by 2):

    6n+3p=546n+3p=54
  4. Multiply first equation by 3:

    3n+3p=363n+3p=36
  5. Subtract this from 6n+3p=546n+3p=54:

    (6n+3p)(3n+3p)=5436(6n+3p)-(3n+3p)=54-36 3n=183n=18 n=6n=6
  6. Substitute into n+p=12n+p=12:

    6+p=12p=66+p=12 \Rightarrow p=6

Answer:

6 notebooks and 6 pens\boxed{6\text{ notebooks and }6\text{ pens}}

Verification:

  • Items: 6+6=126+6=12
  • Cost: 3(6)+1.5(6)=18+9=273(6)+1.5(6)=18+9=27

Both conditions match.

Worked Example 9: No solution / infinite solutions

Worked Example 9: Values of k for no solution and infinite solutions

Given system:

{kx+6y=122x+3y=6\begin{cases} kx+6y=12\\ 2x+3y=6 \end{cases}

First, rewrite each equation in slope-intercept form.

  1. First equation:

    6y=kx+126y=-kx+12 y=k6x+2y=-\frac{k}{6}x+2

    Slope is k6-\frac{k}{6}, intercept is 22.

  2. Second equation:

    3y=2x+63y=-2x+6 y=23x+2y=-\frac{2}{3}x+2

    Slope is 23-\frac{2}{3}, intercept is 22.

For infinite solutions, lines must be identical: same slope and same intercept.

  1. Set slopes equal: k6=23-\frac{k}{6}=-\frac{2}{3} k6=23k=4\frac{k}{6}=\frac{2}{3}\Rightarrow k=4

So infinite solutions when k=4\boxed{k=4}.

For no solution, lines must be parallel but distinct: same slope, different intercept.

  1. Here both intercepts are fixed at 22 regardless of kk.
  2. If slopes are equal, intercepts are also equal, so lines are identical, not distinct.

Therefore, there is no value of kk that gives no solution.

Final:

  • Infinite solutions: k=4\boxed{k=4}
  • No solution: no such k\boxed{\text{no such }k}

Strategy tip: When a parameter like kk appears, convert to slope-intercept form early. It makes solution-type questions much faster.


5. Linear Inequalities

Concept explanation

A linear inequality is solved like a linear equation, with one key rule:

  • If you multiply or divide both sides by a negative number, reverse the inequality sign.

Examples:

  • If 2x>8-2x > 8, then x<4x < -4 (sign flips).
  • If 3x123x \le 12, then x4x \le 4 (no flip, since dividing by positive 3).

Compound inequalities

Compound inequalities use two parts:

  • AND: value must satisfy both conditions (intersection).
  • OR: value can satisfy either condition (union).

Example AND form:

1x<4-1 \le x < 4

Example OR form:

x<2  or  x5x< -2 \;\text{or}\; x\ge 5

Graphing on number lines

  • Open circle for << or >>.
  • Closed circle for \le or \ge.
  • Shade left for "less than," right for "greater than."

Systems of inequalities

On a coordinate plane, each inequality represents a half-plane. The solution region is where all shading overlaps.

In SAT word problems, inequalities often express constraints like "at least," "no more than," "cannot exceed," or "minimum required." Translate those words carefully.

Worked Example 10: Compound inequality

Worked Example 10: Solve -3 ≤ 2x + 5 < 11

Solve:

32x+5<11-3 \le 2x+5 < 11
  1. Subtract 5 from all three parts:

    82x<6-8 \le 2x < 6
  2. Divide all parts by 2 (positive, so no sign flip):

    4x<3-4 \le x < 3

Answer:

4x<3\boxed{-4 \le x < 3}

Interval notation:

[4,3)\boxed{[-4,\,3)}

Worked Example 11: Word problem inequality

Worked Example 11: Minimum score needed

A student needs at least 480 points total on 6 tests. After 5 tests, they have 394 points. What minimum score is needed on the 6th test?

  1. Let xx be the score on test 6.

  2. Write inequality:

    394+x480394 + x \ge 480
  3. Subtract 394:

    x86x \ge 86

Minimum required score:

86\boxed{86}

Interpretation: any score of 86 or higher reaches the goal.

Strategy tip: Words like "at least" map to \ge and "at most" map to \le. Underline those phrases before writing the inequality.


Quick Reference: Key Formulas

Algebra Formula Box: Must-Know SAT Relationships

Slope formula

m=y2y1x2x1m = \frac{y_2-y_1}{x_2-x_1}

Slope-intercept form

y=mx+by = mx + b
  • mm = slope (rate of change)
  • bb = yy-intercept (value when x=0x=0)

Point-slope form

yy1=m(xx1)y - y_1 = m(x - x_1)

Standard form

Ax+By=CAx + By = C

System solution conditions (for lines)

  • One solution: different slopes (m1m2m_1 \neq m_2)
  • No solution: same slope, different intercepts (m1=m2m_1=m_2, b1b2b_1\neq b_2)
  • Infinite solutions: same slope, same intercept (m1=m2m_1=m_2, b1=b2b_1=b_2)

Inequality rules

  • Add/subtract the same quantity on both sides: sign stays.
  • Multiply/divide by positive: sign stays.
  • Multiply/divide by negative: flip inequality sign.

Digital SAT Tips for Algebra

The Digital SAT format makes strategy especially important. Algebra is not just about solving; it is about solving efficiently while choosing the best tool (mental math, algebraic steps, or Desmos).

Desmos and visual checks

Use the built-in Desmos graphing calculator when visual confirmation is faster than symbolic manipulation.

  • For systems: graph both lines and identify the intersection.
  • For equation checks: graph left side and right side as separate expressions; intersection xx gives solution.
  • For inequality boundaries: graph the boundary line and reason about the shaded side.

High-yield moves

  1. For "which equation represents..." questions, match slope and intercept to context first.
  2. Plug in x=0x=0 to find intercept quickly when comparing equivalent equations.
  3. If question asks for an expression like 3x+23x+2, try deriving that expression directly from the given equation instead of solving fully for xx first.
  4. In long equations, simplify both sides independently before moving terms.
  5. In word problems, define variables with units before writing equations.

Common error checklist before submitting

  • Did you distribute negatives correctly?
  • Did you use the correct inequality symbol direction?
  • Did you clear denominators using the LCD on all terms?
  • Did you answer what was asked (for example, 3x+23x+2 instead of xx)?
  • Is your answer reasonable in context (no negative number of items, realistic units)?

Strategy tip: Spend 5 seconds on a reasonableness check. That habit catches many avoidable misses.


Practice Problems

Try these without looking at the answers first.

Problem 1 (Linear equation in one variable)

Solve for xx:

5(2x1)3=4x+165(2x-1)-3 = 4x+16

A) x=1x=1
B) x=2x=2
C) x=3x=3
D) x=4x=4

Problem 2 (Linear equation in two variables / slope)

What is the slope of the line through (2,1)(2, -1) and (8,11)(8, 11)?

A) 11
B) 22
C) 12\frac{1}{2}
D) 2-2

Problem 3 (Linear function from table)

A function is linear and passes through the points (0,4)(0,4) and (5,19)(5,19). Which equation represents the function?

A) f(x)=3x+4f(x)=3x+4
B) f(x)=155x+19f(x)=\frac{15}{5}x+19
C) f(x)=4x+3f(x)=4x+3
D) f(x)=2x+4f(x)=2x+4

Problem 4 (System word problem)

At a fundraiser, adult tickets cost $12 and student tickets cost $8. A total of 40 tickets are sold for $400. How many student tickets were sold?

A) 10
B) 20
C) 25
D) 30

Problem 5 (Linear inequality)

Solve:

4<3x114-4 < 3x - 1 \le 14

A) 1<x5-1 < x \le 5
B) 43<x5-\frac{4}{3} < x \le 5
C) 1<x<5-1 < x < 5
D) 1x5-1 \le x \le 5

Answers and Explanations

1) Correct answer: D (x=4x=4)
Expand and solve:

10x53=4x+1610x8=4x+1610x-5-3=4x+16 \Rightarrow 10x-8=4x+166x=24x=46x=24 \Rightarrow x=4

2) Correct answer: B (m=2m=2)
Use slope formula:

m=11(1)82=126=2m=\frac{11-(-1)}{8-2}=\frac{12}{6}=2

3) Correct answer: A (f(x)=3x+4f(x)=3x+4)
Slope:

m=19450=155=3m=\frac{19-4}{5-0}=\frac{15}{5}=3

Since point (0,4)(0,4) is on the line, intercept is 44. So f(x)=3x+4f(x)=3x+4.

4) Correct answer: B (20 student tickets)
Let aa = adult tickets, ss = student tickets.

a+s=40a+s=4012a+8s=40012a+8s=400

Substitute a=40sa=40-s:

12(40s)+8s=40048012s+8s=4004s=80s=2012(40-s)+8s=400 \Rightarrow 480-12s+8s=400 \Rightarrow -4s=-80 \Rightarrow s=20

5) Correct answer: A (1<x5-1 < x \le 5)
Solve chain inequality:

4<3x114-4<3x-1\le14

Add 1:

3<3x15-3<3x\le15

Divide by 3:

1<x5-1<x\le5

Algebra success on the SAT comes from repetition with structure: set up cleanly, solve step by step, and verify quickly. If you master the patterns in this guide, you will be prepared for the highest-weight Math domain and build momentum for your total score.